Respuesta :
Answer : The % of (+) limonene isomer = 79%
The % of (-) limonene isomer = 0%
The % of enantiomeric excess = 58%
Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.
Given,
% of pure limonene enantiomer = The % of (+) limonene isomer = 79%
Therefore, The % of (-) limonene isomer = 0%
Formula used :
[tex]\%(+)\text{ isomer}=\frac{ee}{2}+50\%[/tex]
Where, ee → enantiomeric excess
Now, put all the values in above formula, we get the value of enantiomeric excess (ee).
[tex]{ee}=\frac{\%(+)-50\%}{2}[/tex]
[tex]=\frac{79\%-50\%}{2}[/tex]
= 58%
Answer:
The major enatiomer makes up 89.5% of the mixture
The minor enatiomer makes up 10.5% of the mixture
The percent enantiometric excess is 79%
Explanation:
% (+) = ee/2
79 % / 2 + 50%
= 39.5% + 50%
= 89.5%
and
% (-) = 100 % - 89.5 %
= 10.5%
